You Try It #5 a. To begin, input the first 130 mg for the 6 a.m. coffee on the homescreen. Now calculate the amount of coffee still present after the second (7 a.m.) cup of coffee. And after the third cup. And after the fourth cup. n (cups) t (hours) f(t) mg 0 130 2 1 243.1 3 2 341.5 4 3 427.1 b. Did you find yourself repeating the same process over and over? Write down what you did at each stage. The calculator commands as followed 130 ENTER ANS(0.87)+130 ENTER ENTER ENTER As shown above represents from the initial intake of 130mg worth of coffee and the addition of the same amount per hour adding on to the prior amount still within the person's body. You Try It #6 a. Solve the pair of equations A + C = 130 and 0.87A + C = 243.1 for A and C. Use your results to write the function f(t) = Art + C that models the amount of caffeine in the body of the person who drank one cup of coffee each hour, each cup containing 130 mg of caffeine, with the body eliminating 13% of the caffeine present each hour. 0.87A+130-A=243.1 - -.13A=113.1 - A=-870-870+C=130 - C=1000 0.874 +1000 f (t)=-870% b. Evaluate your function at t = 2 and at t = 3. f(2) should give you the amount of caffeine in the body 2 hours after he started drinking coffee, thus, just after he drank his third cup. Does your value for f(2) agree with the amount of caffeine you calculated should be present after the third cup in YTI#5? 0.874+1000=341.5 (2)=-8704 , this agrees with the result in YTI#5 c. Evaluate f(3). Compare the result to the corresponding result in YTI#5 and explain what this result tells us. 0.87 4 +1000=427.1 f (3)=-8704 ,agreeing with the corresponding result in YTI#5. f(3) = 427.1 shows that 3 hours after the 1st cup of coffee right after the person drink his/her 4th cup there is about 427 mg of caffeine in their body. You Try It #7 a. Graph the function you developed in YTI#6. Find a window that gives you a good view of its long term behavior. + + A(t) = -870(0.87)' + 1000 b. After studying the graph, describe what eventually happens to the amount of caffeine in a person's body if he continues drinking 1 cup of coffee each hour, not just for two days, but indefinitely, day after day after day. Based on the information if the person continues drinking 1 cup of coffee each our indefinitely, the caffeine in the body approaches an upper boundary of 1000 mg. c. Will a person who drinks a cup of coffee every hour, with 130 mg in each cup, eventually have 5000 mg of caffeine in his/her body? With the amount the person is intaking per hour, they will not reach 5000 mg of caffeine in which their caffeine level will not exceed over 1000 mg. 3. Conclusion Based on the product you can tell how much caffeine a person intakes and how harmful it can be when constantly consuming it. When calculating my own personal consumption of caffeine I realised that I took in a lot throughout the day. With the calculations used to determine how much caffeine was present in my body after a few hours shocked me how long it takes for caffeine to get out of my system to the point that it doesn't act as a minor stimulant. With only 13% of caffeine present being eliminated per hour, it makes me realize how dependent I am with consuming caffeine in order for me to be more awakeand have energy. I enjoyed doing this project because not only it displayed how much I personally intake, but also the effects if I were to constantly consuming coffee hourly to the amount of caffeine is still in a person's system after intaking a certain amount. Caffeine is something that several people consume on a daily basis without truly knowing how much they intake and the effects depending how much is present in their body. F'trarion by the kidneys J5 one of the major proeesses by which our bodies eliminate Cf'l'C'flll'E. The kidneys usually tter a tired fraction of the enemieai present ourtrtg' each hour (or each day. or eachyear}. Caffeine tea a. and cadmium are all eilmtnateo this way. For caem. we measure tin hours because caeme ts ei'ml'nareci from the body fairly quickly For least. we wouto' measure I in days. Cactottunt (a chemical in eigarertesj r's eilmtnated so slot-my that we eouict measure tin years when we merrierI cadmium allot-matron. Another process delimitation is metabolism by enzymes item the ilver. This method often results in a nearly constant amount ofthe chem-teat been; broken down in a given rune pmoo'. instead ofa r'o'eo' fraction 0th as above. Areohot is eliminated 'ont our bodies this way. I.-'.te eanno! use an exponential function to model citmrrtarion of ateoitoi- a rational I'urtcrtoo wot-rs weir for that model See Consortium to fora study or such a model. the discrete model might be written this way.- are) - 130 cm} . t}.8I-'C{n 7.1 . on is the about r-t-t'tar you did mthyoav catuiatol'? What was C(n Ll toy-our calculator computation? iFyou otto'n'r use that. could you have shortened your eaicuia tor thrown by USU'JQ .Ir instead of what you did? We used 0.3mm} beeause tr .5 shorter man C(n} L1 136m}. these computations can aiso be done using a spreadsheet to wmeh the fomwia in eeit A2 below is cooled down as far as necessary. A t 130 2 H.8i'Ah 136' 3 0.3? 162+ rim 1 0.3?'A3+ tJ'G' Hli'L-TIIF Pull-GU? Section: Spring.-".Summer EDGE You have seen by now that a process that eliminates a xed fraction of a chemical from our bodies each hour can be modeled by an exponential function of the form fit} = rlr' . Here t is a length of time since the start A is the amount of the chemical present at the start. and r represents the fraction of the chemical that remains in our body [the fraction that is at eliminated during the hour}. You probably used a piecewise-dened function with 'pieces" involving the form .1111 to model your own caffeine intake and retention. Suppose someone drank one cup of coffee every hour all day long. This might be unrealistic where coffee drinking is concerned. but the approach we develop in this model applies to medicines that are taken every 4 hours. for example. and to cadmium in people who smoke every day. or toxic chemicals in an industrial environment. and much more. 53 imagine. please. a person involved in a high-risk situation who needs to stay awake and working for 43 hours who compulsively gulps down one cup of coffee every hour on the hour beginning at IE am. As before. we assume that each cup contains [30 mg of caffeine. and that 13% of the coffee present in his body at the start of each hour is eliminated during the hour. Explore the dynamics of the caffeine using a graphing calculator. a. To begin. input the rst 130 mg for the E a. m. coffee on the homesereen. Now calculate the amount of coffee still present after the second E? a.m.] cup of ooffce. And after the third cup. And after the fourth cup. b. Did you nd yourself repeating the same process over and over? Write down what you did at each stage. A graphing calculator [or computer spreadsheet program} can do this repetitive work for you. Start again. Input [3|] as before. and ENTER. Use the Answer key {ANS} to represent the amount that was present just after drinking the last cup of colfee. and input a formula that will repeat the process of calculation that you did at each stage. Then just hit ENTER repeatedly to nd out how much caffeine was present after 2. 3. and 11 cups of coffee {that is. at ? a.m.. 8 a.m.. and t} a.m.]. [f you wanted. you could use this prooess to nd out how much caffeine is there at the end of every hour right through this sleepless knight's 43" hour. This process of iteration of a function. using your most recent output as input at the next stage. gives us a discrete model ofthe process that is occurring in the person's body. It allows us to observe the process in steps. But let's try to write a "closed form" function for the evetyrhour-on-the-hour coffee drinker. a function more like the one we used for the caeine drinkers in the rst part of our article. t} = Ad. The big difference between this situation and the one we modeled before is the addition ofa new 13E] mg of coffee each hour. Perhaps. then. the new situation could be modeled by fit} = Fir" . C. Because 13% of the caffeine is eliminated each hour. and thus Brit: is retained. we know that the value of r in this model should be 0.8T. just as before. But. perhaps surprisingly. in this new situation. A is not the initial amount of caffeine. from the rst cup. and Cis not the amount added with each new cup. either. The form of the function is right. but we have to calculate A and If from the basic premises of the situation.